--- title: "Throttled reheat cycle with 50 bar steam" subject: "Power Plant Equipment" chapter: 7 key: "semester_7/power_plant_equipment:throttled-reheat-50-bar" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/throttled-reheat-50-bar site: BCT Engineering --- # Throttled reheat cycle with 50 bar steam Power Plant Equipment · Chapter 7: Thermal (Steam) Power Plant - Asked in: 2076 Asoj, 10 marks - Video: [Thermodynamics - IDEAL REHEAT RANKINE CYCLE](https://www.youtube.com/watch?v=WTDW_AqqY-I) — JG ENGINEERING TUTORIALS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/throttled-reheat-50-bar - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-7 ## Question Steam is generated in a boiler at 50 bar and 450°C. For the purpose of governing, the steam is throttled to 30 bar before it enters the high pressure stage of turbine. After expansion in high pressure stage, the steam emerges just dry saturated and then reheated at the same pressure to 300°C before it expanded in the low pressure stage to a pressure of 0.06 bar, when it emerges again just dry and saturated. If the intermediate pressure is 3 bar, what are the stage efficiencies? Also calculate the overall cycle efficiency and work ratio shared by HP stage to that of work shared by LP stage. [Superheated and saturated steam tables were attached to the paper] ## Short answer Based on stop-valve state for HP, η_HP = 89.5%; LP stage η_LP = 72.2%. Overall cycle efficiency is 31.1% (or 31.2% neglecting pump work), and the HP to LP work ratio is 1.18. ## Answer **Data:** boiler 50 bar, 450 C; throttled to 30 bar; HP stage 30 bar to 3 bar, exit dry saturated; reheat at 3 bar to 300 C; LP stage 3 bar to 0.06 bar, exit dry saturated. Values from steam tables (IAPWS-IF97). | State | p | Condition | h (kJ/kg) | s (kJ/kg K) | |---|---|---|---|---| | 1 | 50 bar | 450 C | 3317.0 | 6.8208 | | 1' | 30 bar | after throttling, $h = h_1$ (about 437.7 C) | 3317.0 | 7.0468 | | 2 | 3 bar | dry saturated (actual HP exit) | 2724.9 | 6.9916 | | 3 | 3 bar | 300 C | 3069.6 | 7.7037 | | 4 | 0.06 bar | dry saturated (actual LP exit) | 2566.7 | 8.3291 | | 5 | 0.06 bar | saturated water | 151.49 | 0.5209 | At 3 bar: $h_f = 561.46$, $h_{fg} = 2163.44$, $s_f = 1.6718$, $s_{fg} = 5.3198$. At 0.06 bar: $h_{fg} = 2415.17$, $s_{fg} = 7.8083$, $v_f = 0.001006$. ```text 1 (50 bar) --throttle--> 1' (30 bar) --HP stage--> 2 (3 bar) --reheat--> 3 (3 bar) --LP stage--> 4 (0.06 bar) --condenser--> 5 --pump--> boiler ``` #### HP stage efficiency Throttling is the governing device of the HP stage, so the isentropic reference is taken from the stop-valve state 1 (50 bar, 450 C) to 3 bar: $$ \begin{aligned} x_{2s} &= \frac{6.8208 - 1.6718}{5.3198} = 0.9679 \\ h_{2s} &= 561.46 + 0.9679 \times 2163.44 = 2655.4\ \text{kJ/kg} \\ \eta_{HP} &= \frac{h_1 - h_2}{h_1 - h_{2s}} = \frac{3317.0 - 2724.9}{3317.0 - 2655.4} = \frac{592.1}{661.6} = 0.895 \end{aligned} $$ Note: if the reference is taken from the throttled state 1' ($s = 7.0468 > s_g$ at 3 bar), the isentropic end point is superheated with $h = 2747.6$ kJ/kg, which is higher than the given actual exit (2724.9). That would give an efficiency above 100%, so with these data the stage efficiency must be based on the stop-valve state (it then includes the throttling loss). #### LP stage efficiency $$ \begin{aligned} x_{4s} &= \frac{7.7037 - 0.5209}{7.8083} = 0.9199 \\ h_{4s} &= 151.49 + 0.9199 \times 2415.17 = 2373.2\ \text{kJ/kg} \\ \eta_{LP} &= \frac{h_3 - h_4}{h_3 - h_{4s}} = \frac{3069.6 - 2566.7}{3069.6 - 2373.2} = \frac{502.9}{696.4} = 0.722 \end{aligned} $$ #### Overall cycle efficiency $$ \begin{aligned} W_{HP} &= 3317.0 - 2724.9 = 592.1\ \text{kJ/kg} \\ W_{LP} &= 3069.6 - 2566.7 = 502.9\ \text{kJ/kg} \\ W_p &= 0.001006 (5000 - 6) = 5.03\ \text{kJ/kg} \\ Q_s &= (h_1 - h_5 - W_p) + (h_3 - h_2) \\ &= (3317.0 - 151.49 - 5.03) + (3069.6 - 2724.9) \\ &= 3160.5 + 344.7 = 3505.2\ \text{kJ/kg} \\ \eta &= \frac{592.1 + 502.9 - 5.03}{3505.2} = \frac{1090.0}{3505.2} = 0.311 \end{aligned} $$ (Neglecting pump work: $\eta = 1095.0 / 3510.3 = 0.312$.) #### Work ratio HP : LP $$ \frac{W_{HP}}{W_{LP}} = \frac{592.1}{502.9} = 1.177 $$ | Result | Value | |---|---| | HP stage efficiency | 89.5% (from stop valve) | | LP stage efficiency | 72.2% | | Overall cycle efficiency | 31.1% (31.2% without pump work) | | Work shared HP : LP | 1.177 : 1 | **Answer: $\eta_{HP} \approx 89.5\%$, $\eta_{LP} \approx 72.2\%$, overall $\eta \approx 31.1\%$, $W_{HP}/W_{LP} \approx 1.18$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is conserved during throttling from 50 bar to 30 bar? A. Entropy B. Enthalpy (h = h1 = 3317.0 kJ/kg) C. Pressure D. Temperature exactly Answer: B. Throttling is a constant-enthalpy process, so h1' = h1 while entropy increases. (Key point: Throttling process) 2. Why is HP stage efficiency based on the stop-valve state (50 bar, 450 C)? A. From the throttled state the isentropic end would be superheated at higher h than the actual exit, giving efficiency over 100% B. Because pressure is higher there C. Because the pump is included D. Because reheat has not happened Answer: A. Using 30 bar gives h of 2747.6 above the actual 2724.9, so the stage efficiency must be based on state 1. (Key point: HP efficiency reference) 3. HP stage efficiency is approximately: A. 72.2% B. 89.5% C. 31.1% D. 96.8% Answer: B. 592.1/661.6 = 0.895. (Key point: HP stage efficiency) 4. LP stage efficiency = (h3 - h4)/(h3 - h4s) with isentropic dryness 0.9199. Its value is about: A. 89.5% B. 72.2% C. 118% D. 31% Answer: B. 502.9/696.4 = 0.722. (Key point: LP stage efficiency) 5. Overall cycle efficiency and HP:LP work ratio are: A. 31.1% and 1.18 B. 72.2% and 0.85 C. 89.5% and 1.18 D. 31.1% and 0.85 Answer: A. Net work 1090.0 over Q_s 3505.2 gives 31.1%; 592.1/502.9 = 1.18. 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